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prove by mathematical induction that 7^n+4^n+1 is divisible by 6

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asked by JESS
  1. 7^n + 4^n + 1

    1. test for n = 1
    7^1 + 4^1 + 1 = 12 , which is divisible by 6

    2. assume it is true for n = k , that is, assume that
    7^k + 4^k + 1 is divisible by 6

    3. then show that 7^(k+1) + 4^(k+1) - 1 is divisible by 6

    use the number property that if both A and B are divisble by c
    then A-B is divisible by c
    e.g. 156 and 117 are both divisible by 13
    then is 156-117 or 39 divisible by 13 ? YES

    so ...
    7^(k+1) + 4^(k+1) + 1 - (7^k + 4^k + 1)
    = 7^(k+1) + 4^(k+1) + 1 -7^k - 4^k - 1
    = 7^k(7-1) + 4^k(4-1)
    = 6(7^k) + 3(4^k)

    clearl 6(7^k) is a multiple of 6 , thus divisible by 6
    and in 3(4^k) , the 4^k must be even and any even times 3 is divisible by 6
    so we have shown that the result is divisible by 6
    (the sum of multiples of 6 must be divisible by 6)

    posted by Reiny

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